A glass full of water has a bottom of area `20 cm^(2)`, top of area `20cm^(2)`, height `20 cm` and volume half a litre.

a. Find the force exerted by the water on the bottom.
b. Considering the equilibrium of the water. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure `= 1.0 xx 10^(5) N//m^(2)`. Density of water `= 1000 kg//m^(-3)` and `g = 10 m//s^(2)`

Answer

A, B, D

Explanation

`Pa=1.0xx10^5N/m^2`,
`=Pw=10^3mg/m^3`
g=10m/s^2`
`v=500m/=500g=0.5kg`
a. Force exerted at the bottom = Force due to cylindrical
water column + atm. Force
`=AxxhxxP_wxxg+P_axxA`
`=A(hrho_wg+P_a)`
`=204N`
b. To find out the resultant force exerted by the sides of the glass, from the free body diagram of water inside the glass,
`P_axxA+mg=Axxhxxrho_2xxg+F_s+P_axxA`
`rarr mg=Axxhxxp_wxg+F_s`
`rarr 0.5xx1=20xx10^-4xx20x10^-2xx10^-3xx10+F_s`
`rarr F_s=5-4=1N` (upward)
This force 1N is provided by the sides of the glass.